$f \in (\mathbb{R}^{3})^*$ is a functional of the form $f(x_1, x_2, x_3) = x_1 - 3x_2 + x_3$.
$\mathcal{A} = \{α_1, α_2, α_3\}$ is a basis of the linear space $\mathbb{R}^{3}$ and $\mathcal{A}^* = \{g_1, g_2, g_3\} ⊂ (\mathbb{R}^{3})^*$ is a dual basis.
- $g_1(x_1, x_2, x_3) = x_1 - x_3$
- $g_2(x_1, x_2, x_3) = x_1 - 2x_2$
Assuming that the third coordinate of the functional $f$ in the basis $\mathcal{A}^∗$ is equal to $1$, find the vector $α_3$.
So I don't know $g_3$ but I know that "the third coordinate of the function $f$ in the basis $\mathcal{A}^∗$ is equal to $1$". I think that I need to use that information to get full $\mathcal{A}^∗$, but I'm not sure how. My first idea was to express $g_3(x_1, x_2, x_3) = ax_1 + bx_2 + cx_3$ in matrix:
\begin{bmatrix} 1 & 1 & a & 1\\ 0 & -2 & b & -3\\ -1 & 0 & c & 1 \end{bmatrix}
with an aim to get:
\begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & ?\\ 0 & 0 & 1 & ? \end{bmatrix}
And then somehow find those $3$ variables. But that does not work for me and I don't know what to do next.
Are you sure the dual basis isn't just an arbitrary basis but the one induced by the basis of $\mathbb{R}^3$? Your language of "a basis" is unusual; it should probably be "the corresponding basis." Once you pick a basis of $V$ you get a naturally induced basis of $V^*$.
If you do assume the $g_i$ are induced by the $\alpha_i$, then $g_i(\alpha_j)=\delta_{ij}$, the Kronecker delta (which is $1$ when $i=j$ and $0$ otherwise).
Thus $g_1(\alpha_3)=0$, $g_2(\alpha_3)=0$, and $g_3(\alpha_3)=1$.
Assume $\alpha_3=(a,b,c)$.
The first equation above tells you $a=c$, so $\alpha_3=(a,b,a)$.
The second equation tells you $a=2b$, so $\alpha_3=(2b, b, 2b)$.
Now, we also know that the third coordinate of $f$ with respect to the $g_i$ is 1, so
$$f=c_1g_1+c_2g_2+g_3$$
Thus
$$2b-3b+2b=f(\alpha_3)=c_1\underbrace{g_1(\alpha_3)}_{\text{zero}}+c_2\underbrace{g_2(\alpha_3)}_{{\text{zero}}}+\underbrace{g_3(\alpha_3)}_{\text{one}}=1$$
So $b=1$.
Thus $\alpha_3=(2,1,2)$.